[R] colnames
Gilbert Wu
gilbert.wu at sabrefund.com
Thu Jul 21 18:56:55 CEST 2005
Hi Adai,
Your diagnosis is absolutely right; class(r1) returned data.frame and your suggested solution worked perfectly. Your assumption is also right; both x and y are positive.
If I want to compare the performance of the my old function with yours, are there some functions in R I could use to get the elapsed time etc?
Many Thanks indeed.
Regards,
Gilbert
-----Original Message-----
From: Adaikalavan Ramasamy [mailto:ramasamy at cancer.org.uk]
Sent: 19 July 2005 23:38
To: Gilbert Wu
Cc: r-help at stat.math.ethz.ch
Subject: RE: [R] colnames
What does class(r1) give you ? If it is "data.frame", then try
exp( diff( log( as.matrix( df ) ) ) )
BTW, I made the assumption that both x and y are positive values only.
Regards, Adai
On Tue, 2005-07-19 at 16:30 +0100, Gilbert Wu wrote:
> Hi Adai,
>
> When I tried the optimized routine, I got the following error message:
>
> r1
> 899188 902232 901714 28176U 15322M
> 20050713 7.595 10.97 17.96999 5.1925 11.44
> 20050714 7.605 10.94 18.00999 5.2500 11.50
> 20050715 7.480 10.99 17.64999 5.2500 11.33
> 20050718 7.415 11.05 17.64000 5.2250 11.27
> > exp(diff(log(r1))) -1
> Error in r[i1] - r[-length(r):-(length(r) - lag + 1)] :
> non-numeric argument to binary operator
> >
>
> Any idea?
>
> Many Thanks.
>
> Gilbert
> -----Original Message-----
> From: Adaikalavan Ramasamy [mailto:ramasamy at cancer.org.uk]
> Sent: 19 July 2005 12:20
> To: Gilbert Wu
> Cc: r-help at stat.math.ethz.ch
> Subject: RE: [R] colnames
>
>
> First, your problem could be boiled down to the following example. See
> how the colnames of the two outputs vary.
>
> df <- cbind.data.frame( "100"=1:2, "200"=3:4 )
> df/df
> X100 X200
> 1 1 1
> 2 1 1
>
> m <- as.matrix( df ) # coerce to matrix class
> m/m
> 100 200
> 1 1 1
> 2 1 1
>
> It appears that whenever R has to create a new dataframe automatically,
> it tries to get nice colnames. See help(data.frame). I am not exactly
> sure why this behaviour is different when creating a matrix. But I do
> not think this is a major problem for most people. If you coerce your
> input to matrix, the problem goes away.
>
>
> Next, note the following points :
> a) "mat[ 1:3, 1:ncol(mat) ]" is equivalent to simply "mat[ 1:3, ]".
> b) "mat[ 2:nrow(mat), ]" is equivalent to simply "mat[ -1, ]"
> See help(subset) for more information.
>
> Using the points above, we can simplify your function as
>
> p.RIs2Returns <- function (mat){
>
> mat <- as.matrix(mat)
> x <- mat[ -nrow(mat), ]
> y <- mat[ -1, ]
>
> return( y/x -1 )
> }
>
> If your data contains only numerical data, it is probably good idea to
> work with matrices as matrix operations are faster.
>
>
> Finally, we can shorten your function. You can use the diff (which works
> column-wise if input is a matrix) and apply function if you know that
>
> y/x = exp(log(y/x)) = exp( log(y) - log(x) )
>
> which could be coded in R as
>
> exp( diff( log(r1) ) )
>
> and then subtract 1 from above to get your returns.
>
> Regards, Adai
>
>
>
> On Tue, 2005-07-19 at 09:17 +0100, Gilbert Wu wrote:
> > Hi Adai,
> >
> > Many Thanks for the examples.
> >
> > I work for a financial institution. We are exploring R as a tool to implement our portfolio optimization strategies. Hence, R is still a new language to us.
> >
> > The script I wrote tried to make a returns matrix from the daily return indices extracted from a SQL database. Please find below the output that produces the 'X' prefix in the colnames. The reason to preserve the column names is that they are stock identifiers which are to be used by other sub systems rather than R.
> >
> > I would welcome any suggestion to improve the script.
> >
> >
> > Regards,
> >
> > Gilbert
> >
> > > "p.RIs2Returns" <-
> > + function (RIm)
> > + {
> > + x<-RIm[1:(nrow(RIm)-1), 1:ncol(RIm)]
> > + y<-RIm[2:nrow(RIm), 1:ncol(RIm)]
> > + RReturns <- (y/x -1)
> > + RReturns
> > + }
> > >
> > >
> > > channel<-odbcConnect("ourSQLDB")
> > > result<-sqlQuery(channel,paste("select * from equityRIs;"))
> > > odbcClose(channel)
> > > result
> > stockid sdate dbPrice
> > 1 899188 20050713 7.59500
> > 2 899188 20050714 7.60500
> > 3 899188 20050715 7.48000
> > 4 899188 20050718 7.41500
> > 5 902232 20050713 10.97000
> > 6 902232 20050714 10.94000
> > 7 902232 20050715 10.99000
> > 8 902232 20050718 11.05000
> > 9 901714 20050713 17.96999
> > 10 901714 20050714 18.00999
> > 11 901714 20050715 17.64999
> > 12 901714 20050718 17.64000
> > 13 28176U 20050713 5.19250
> > 14 28176U 20050714 5.25000
> > 15 28176U 20050715 5.25000
> > 16 28176U 20050718 5.22500
> > 17 15322M 20050713 11.44000
> > 18 15322M 20050714 11.50000
> > 19 15322M 20050715 11.33000
> > 20 15322M 20050718 11.27000
> > > r1<-reshape(result, timevar="stockid", idvar="sdate", direction="wide")
> > > r1
> > sdate dbPrice.899188 dbPrice.902232 dbPrice.901714 dbPrice.28176U dbPrice.15322M
> > 1 20050713 7.595 10.97 17.96999 5.1925 11.44
> > 2 20050714 7.605 10.94 18.00999 5.2500 11.50
> > 3 20050715 7.480 10.99 17.64999 5.2500 11.33
> > 4 20050718 7.415 11.05 17.64000 5.2250 11.27
> > > #Set sdate as the rownames
> > > rownames(r1) <-as.character(r1[1:nrow(r1),1:1])
> > > #Get rid of the first column
> > > r1 <- r1[1:nrow(r1),2:ncol(r1)]
> > > r1
> > dbPrice.899188 dbPrice.902232 dbPrice.901714 dbPrice.28176U dbPrice.15322M
> > 20050713 7.595 10.97 17.96999 5.1925 11.44
> > 20050714 7.605 10.94 18.00999 5.2500 11.50
> > 20050715 7.480 10.99 17.64999 5.2500 11.33
> > 20050718 7.415 11.05 17.64000 5.2250 11.27
> > > colnames(r1) <- as.character(sub("[[:alnum:]]*\\.","", colnames(r1)))
> > > r1
> > 899188 902232 901714 28176U 15322M
> > 20050713 7.595 10.97 17.96999 5.1925 11.44
> > 20050714 7.605 10.94 18.00999 5.2500 11.50
> > 20050715 7.480 10.99 17.64999 5.2500 11.33
> > 20050718 7.415 11.05 17.64000 5.2250 11.27
> > > RRs<-p.RIs2Returns(r1)
> > > RRs
> > X899188 X902232 X901714 X28176U X15322M
> > 20050714 0.001316656 -0.002734731 0.002225933 0.011073664 0.005244755
> > 20050715 -0.016436555 0.004570384 -0.019988906 0.000000000 -0.014782609
> > 20050718 -0.008689840 0.005459509 -0.000566006 -0.004761905 -0.005295675
> > >
> >
>
>
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